Sampling Distribution of a Sample Mean Homework 1. Suppose that the number of texts sent during a typical day by a randomly selected high school student follows a right-skewed distribution with a mean of 15 and a standard deviation of 35 . Assuming that students at your school are typical texters, how likely is it that a random sample of 50 students will have sent more than a total of 1000 texts in the last 24 hours? 2. The number of absences last year for students at a large high school has a mean of 5.6 days with a standard deviation of 4.1 days. Suppose we take an SRS of 80 of last year's students and calculate the mean number of absences for the members of the sample. (Refer to example of page 470 in your textbook) (a) Identify the mean of the sampling distribution of $$ \bar{x} $$. (b) Calculate and interpret the standard deviation of the sampling distribution of $$ \bar{x} $$. Verify that the $$ 10 \% $$ condition is met.

Question

UpStudy AI · Accepted Answer

Sure, let's tackle each of your homework questions step by step.

---

## **1. Probability of Sending More Than 1000 Texts by 50 Students**

**Given:**
- **Mean ($$ \mu $$)**: 15 texts per student per day
- **Standard Deviation ($$ \sigma $$)**: 35 texts per student per day
- **Sample Size ($$ n $$)**: 50 students
- **Total Texts ($$ S $$)**: 1000 texts

**Objective:**
Find the probability that the total number of texts sent by 50 students exceeds 1000 in a day.

**Approach:**
We can use the **Central Limit Theorem (CLT)**, which states that the sampling distribution of the sum (or average) of a large number of independent, identically distributed variables will be approximately normally distributed, regardless of the original distribution's shape.

1. **Calculate the Mean of the Sum ($$ \mu_S $$):**
   $$
   \mu_S = n \times \mu = 50 \times 15 = 750 \text{ texts}
   $$

2. **Calculate the Standard Deviation of the Sum ($$ \sigma_S $$):**
   $$
   \sigma_S = \sqrt{n} \times \sigma = \sqrt{50} \times 35 \approx 7.071 \times 35 \approx 247.49 \text{ texts}
   $$

3. **Find the Z-Score:**
   $$
   Z = \frac{S - \mu_S}{\sigma_S} = \frac{1000 - 750}{247.49} \approx \frac{250}{247.49} \approx 1.010
   $$

4. **Determine the Probability:**
   Using standard normal distribution tables or a calculator:
   $$
   P(Z > 1.010) \approx 0.156 \text{ or } 15.6\%
   $$

**Conclusion:**
There is approximately a **15.6% probability** that a random sample of 50 students will send more than a total of 1000 texts in the last 24 hours.

---

## **2. Sampling Distribution of the Sample Mean for Student Absences**

**Given:**
- **Population Mean ($$ \mu $$)**: 5.6 days
- **Population Standard Deviation ($$ \sigma $$)**: 4.1 days
- **Sample Size ($$ n $$)**: 80 students

### **(a) Mean of the Sampling Distribution of $$ \bar{x} $$**

**Definition:**
The mean of the sampling distribution of the sample mean ($$ \mu_{\bar{x}} $$) is equal to the population mean.

**Calculation:**
$$
\mu_{\bar{x}} = \mu = 5.6 \text{ days}
$$

**Conclusion:**
The mean of the sampling distribution of $$ \bar{x} $$ is **5.6 days**.

---

### **(b) Standard Deviation of the Sampling Distribution of $$ \bar{x} $$ and the 10% Condition**

**Calculation of Standard Deviation ($$ \sigma_{\bar{x}} $$):**
$$
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.1}{\sqrt{80}} \approx \frac{4.1}{8.944} \approx 0.458 \text{ days}
$$

**Interpretation:**
The standard deviation of the sampling distribution ($$ \sigma_{\bar{x}} $$) is **approximately 0.458 days**. This means that the sample mean number of absences is expected to vary by about 0.458 days from the true population mean of 5.6 days on average.

**Verifying the 10% Condition:**

**What is the 10% Condition?**
The **10% condition** ensures that the sample size is no more than 10% of the population when sampling without replacement. This condition helps in maintaining the independence of observations, which is an assumption for the CLT.

**Application:**
- **Sample Size ($$ n $$)**: 80
- **Population Size ($$ N $$)**: Not specified, but it's stated as a "large high school."

To satisfy the 10% condition:
$$
n \leq 0.10 \times N \Rightarrow N \geq 10 \times n = 10 \times 80 = 800
$$

**Assumption:**
Given that it's a large high school, it's reasonable to assume that:
$$
N \geq 800
$$
Thus, $$ n = 80 $$ is less than 10% of the population.

**Conclusion:**
**Yes, the 10% condition is met** because the sample size of 80 students is assumed to be less than 10% of the total student population.

---
**1. Probability of Sending More Than 1000 Texts:** There's about a 15.6% chance that 50 students will send more than 1000 texts in a day. **2. Sampling Distribution of Absences:** - **Mean ($$ \mu_{\bar{x}} $$)**: 5.6 days - **Standard Deviation ($$ \sigma_{\bar{x}} $$)**: Approximately 0.458 days - **10% Condition**: Met, as the sample size is less than 10% of the population.

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Statistics Questions